Linear Algebra (basis)

  • #1
BrownianMan
134
0
1. In each case, find a basis of the subspace U:

(a) U=span{[1 -1 2 5 1].[3 1 4 2 7],[1 1 0 0 0],[5 1 6 7 8]}

(b) U=span{[1 5 -6]^T, [2 6 -8]^T, [3 7 -10]^T, [4 8 12]^T}

2. Determine if the following sets of vectors are a basis of the indicated space:

{[1 0 -2 5]^T,[4 4 -3 2]^T,[0 1 0 -3]^T,[1 3 3 -10]^T} in R^4


For 1a I get: {[1 -1 2 5 1],[0 4 -2 -13 4],[0 0 2 -3 6]}.

For 1b I get: {[1 5 -6]^T,[2 6 8]^T,[3 7 -10]^T}

Is this right?

For 2, I know that the answer is no, but I'm not sure how to show it. Any help?
 

Answers and Replies

  • #2
Robert1986
828
2
1. In each case, find a basis of the subspace U:

(a) U=span{[1 -1 2 5 1].[3 1 4 2 7],[1 1 0 0 0],[5 1 6 7 8]}

(b) U=span{[1 5 -6]^T, [2 6 -8]^T, [3 7 -10]^T, [4 8 12]^T}

2. Determine if the following sets of vectors are a basis of the indicated space:

{[1 0 -2 5]^T,[4 4 -3 2]^T,[0 1 0 -3]^T,[1 3 3 -10]^T} in R^4


For 1a I get: {[1 -1 2 5 1],[0 4 -2 -13 4],[0 0 2 -3 6]}.

For 1b I get: {[1 5 -6]^T,[2 6 8]^T,[3 7 -10]^T}

Is this right?

For 2, I know that the answer is no, but I'm not sure how to show it. Any help?
How did you come up with your answers to parts 1a and 1b? Here is how I would attack this: You need to find a linear independent set of vectors whose span is equal to U. Now, you know that the vectors given in both parts span U, by the very definition of U. "All" you need to do is to determine a linearly independent subset of the vectors given in each problem.

So, for example, in 1a, the vector [5,1,6,7,8] is just the sum of the first three, so you can throw that vector out and you are left with the other three vectors. Now, is [1,1,0,0,0] a linear combination of the first two? Well, it certainly doesn't look like it, but I will let you figure it out. Then just do the same with problem 1b.

For problem 2, observe that you are trying to figure out if the 4 vectors given are linearly independent and span R^4. Now, since the dimension of R^4 is 4, then any basis of R^4 is a set of 4 linearly ID vectors, and also any set of 4 linearly ID vectors is a basis for R^4. You are given 4 vectors, now you just need to determine whether or not they are Lin ID. If they are Lin ID, they are a basis, if they are not Lin ID, they are not a basis.
 
  • #3
BrownianMan
134
0
For problem 2, observe that you are trying to figure out if the 4 vectors given are linearly independent and span R^4. Now, since the dimension of R^4 is 4, then any basis of R^4 is a set of 4 linearly ID vectors, and also any set of 4 linearly ID vectors is a basis for R^4. You are given 4 vectors, now you just need to determine whether or not they are Lin ID. If they are Lin ID, they are a basis, if they are not Lin ID, they are not a basis.

That's my problem. I'm not sure how to show that they are not linearly independent. I know that to show they are linearly independent you have to show that the only solution that vanishes is the trivial solution.
 
  • #4
Robert1986
828
2
That's my problem. I'm not sure how to show that they are not linearly independent. I know that to show they are linearly independent you have to show that the only solution that vanishes is the trivial solution.

Gotcha, then you need to solve this equation:

x_1[1 0 -2 5]+x_2[4 4 -3 2]+x_3[0 1 0 -3]+x_4[1 3 3 -10]=[0,0,0,0]

If there is a solution s.t. not all of the x_i's are zero, then the set is lin dep, otherwise it is lin id.
 

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